Final answer:
The correct statement concerning the relationship between the net radiation and view factors of two concentric tubes, S1 and S2, with the same temperature and as diffuse emitters, is option C: The view factor of S1 to S2 will be greater than S2 to S1 because S1, being the larger tube, surrounds S2 and has a larger surface area to intercept radiation from S2.
Step-by-step explanation:
When considering the net radiation transfer between two objects, or in this case tubes S1 and S2, the concept of view factors becomes important. View factors represent the fraction of radiation leaving a surface that is intercepted by another surface. Given that both tubes are of the same temperature and are diffuse emitters, we examine the statements provided.
A. Net Radiation Leaving S2
Since the tubes are at the same temperature, the amount of radiation emitted by S2 will be equal to the radiation absorbed by S2 from S1, assuming no loss to the environment. Therefore, the net radiation leaving S2 will be the same as that reabsorbed into S2, making statement A incorrect.
B. View Factor of S2 to Itself
The view factor of S2 to itself would be zero because it is not possible for the solid tube to intercept its own radiation; whereas, S1, being a hollow tube, will have a view factor to itself that is greater than zero. Thus, statement B is also incorrect.
C. View Factor of S1 to S2
The view factor of S1 to S2 will indeed be higher than that from S2 to S1 because S1, the larger tube, surrounds S2 and has a larger surface area that can view S2. Therefore, statement C is true. The correct answer is C, the view factor of S1 to S2 will be greater than S2 to S1.